To calculate the Pb2+ and IO3- concentrations when the mixture comes to equilibrium, we can use the stoichiometry of the balanced equation and the given initial concentrations.
1. Calculate the number of moles of lead (II) nitrate (Pb(NO3)2):
Pb(NO3)2 = (0.05 L)(0.05 M) = 0.0025 mol
2. Calculate the number of moles of sodium iodate (NaIO3):
NaIO3 = (0.04 L)(0.2 M) = 0.008 mol
3. Determine the limiting reagent:
In this case, NaIO3 is in excess, so it is not limiting the reaction. Pb(NO3)2 is limiting.
4. Calculate the number of moles of Pb(IO3)2 formed:
Pb(IO3)2 formed = 0.0025 mol (1 mol Pb(NO3)2 produces 1 mol Pb(IO3)2)
5. Calculate the remaining moles of NaIO3 after the reaction:
NaIO3 remaining = 0.008 mol - 2(Pb(NO3)2) = 0.008 mol - 2(0.0025 mol) = 0.008 mol - 0.005 mol = 0.003 mol
6. Calculate the concentration of IO3- in the solution:
(IO3-) = 0.003 mol / 0.090 L = 0.0333 M
Therefore, the concentrations of Pb2+ and IO3- in a solution saturated with Pb(IO3)2 and with the excess of 0.003 mol / 0.090 L (or 0.0333 M) of IO3- are 0 M and 0.0333 M, respectively.
Finally, we can use the Ksp expression to check if our equilibrium concentrations are consistent with the given Ksp value:
Ksp = (Pb2+)(IO3-)2 = (0 M)(0.0333 M)2 = 0
Since the Ksp is significantly smaller than the given value (2.6 x 10-13), it indicates that the reaction did not reach equilibrium and more Pb(IO3)2 would need to dissolve to reach saturation.